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Im looking for a simple iteration that computes the cube root of $2$ in base $10$(decimal) or base $16$(hexadecimal).

Probably the simplest way is iterating a small polynomial with simple rational coefficients of the form $\frac{n}{10}$ or $\frac{n}{16}$.

Newton and Householder methods do not provide that I think. And neither do continued fractions I assume. I considered median methods but without succes.

I know this case for the pythagoras constant $\sqrt 2$ in hexadecimal(base $16$):

$$\frac{1}{2}x(3 - \frac{1}{2} x^2)(1 + \frac{3}{32}(x^2 - 2)^2)(1 - \frac{1}{64}(x^2 - 2)^3)(1 + \frac{16+11}{2048}(x^2 - 2)^4)$$

is going to $\sqrt 2$ with speed $\frac{3 x^5}{4}$.

So Im motivated to find the apparantly harder cube root of $2$.

edit

I found a solution but it converges slowly.

Are there solutions that converge faster ??

mick
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    You may find this paper interesting: Bombieri, E., van der Poorten, A.J. (1995). Continued Fractions of Algebraic Numbers. In: Bosma, W., van der Poorten, A. (eds) Computational Algebra and Number Theory. Mathematics and Its Applications, vol 325. Springer, Dordrecht. https://doi.org/10.1007/978-94-017-1108-1_10 – Gerry Myerson Apr 02 '23 at 22:33
  • @GerryMyerson seems quite different not ? As for continued fractions ( kinda ) Im still wondering about this : https://math.stackexchange.com/questions/4664497/simplifying-a-complicated-continued-fraction-expression – mick Apr 02 '23 at 22:36
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    The paper is available at https://web.williams.edu/Mathematics/sjmiller/public_html/book/papers/vdp/BombieriPoorten_CFofAlgNumbs.pdf You want to compute $\root3\of2$; the paper is about computing continued fractions for algebraic numbers, and from the continued fraction, you can get as many decimals (or hexadecimals) as you want. – Gerry Myerson Apr 02 '23 at 22:37
  • see the edit for what I want – mick Apr 05 '23 at 23:16
  • Do you want an iteration where division is only by the base or powers of the base? Do you expect that all the digits in each iteration step are valid? Or that one can give exactly how many digits are exact (if the same rounding is applied to the exact result)? How valid is this restriction, for bignum arithmetic division by small integers is still simple. – Lutz Lehmann Apr 06 '23 at 08:12
  • @LutzLehmann The division is only by the base or powers of the base. And I want fast convergeance. This fast convergeance can be used to approximate the correct amount of digits. – mick Apr 06 '23 at 22:46

1 Answers1

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I found $2$ results.

$$z - \frac{(z^3 - 2)(z+1)}{10}$$

for decimal and

$$z - \frac{(z^3 - 2)(z+1)}{16}$$

for hexadecimal.

In fact they can be ajusted to the base you want to use.

For bases larger than $6$ :

$$z - \frac{(z^3 - 2)(z+1)}{a}$$

for base $a$.

They clearly fail for starting point at $-1$ but for real numbers between $\frac{-3}{4}$ and $2$ they usually work.

the julia set picture :

Julia set for the decimal case

mick
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